2. Protonated Salts

1

• 2. Protonated Salts
• These are usually amphoteric salts which react as
  acids and bases. For example, NaH2PO4 in water
  would show the following equilibria
• H2PO4- D H HPO42-
• H2PO4- H2O D OH- H3PO4
• H2O D H OH-
• Hsolution HH2PO4- HH2O OH-H2PO4-
• Hsolution HPO42- OH- H3PO4

2

• Now make all terms as functions in either H or
  H2PO4-, then we have
• H ka2 H2PO4-/H kw/H
  H2PO4-H/ka1
• Rearrangement gives
• H (ka1kw ka1ka2H2PO4-)/(ka1
  H2PO4-1/2
• At high salt concentration and low ka1 this
  relation may be approximated to
• H ka1ka21/2
• Where the pH will be independent on salt
  concentration but only on the equilibrium
  constants.

3

• Example
• Find the pH of a 0.10 M NaHCO3 solution. ka1
  4.3 x 10-7, ka2 4.8 x 10-1
• Solution
• HCO3- D H CO32-
• HCO3- H2O D OH- H2CO3
• H2O D H OH-
• H (ka1kw ka1ka2HCO3-)/(ka1
  HCO3-1/2
• H (4.3x10-7 10-14 4.3x10-7 4.8x10-11
  0.10)/(4.3x10-7 0.10)1/2
• H 4.5x10-9 M
• pH 8.34

4

• The same result can be obtained if we use
• H ka1ka21/2
• H 4.3x10-7 4.8x10-111/2 4.5x10-9 M
• This is since the salt concentration is high
  enough. Now look at the following example and
  compare

5

• Example
• Find the pH of a 1.0x10-4 M NaHCO3 solution. ka1
  4.3 x 10-7, ka2 4.8 x 10-11
• Solution
• H (4.3x10-7 10-14 4.3x10-7 4.8x10-11
  1.0x10-4)/(4.3x10-7 1.0x10-4)1/2
• H 7.97x10-9 M
• pH 8.10
• Substitution in the relation H ka1ka21/2
  will give
• H 4.3x10-7 4.8x10-111/2 4.5x10-9 M,
  which is incorrect
• You can see the difference between the two
  results.

6

• Protonated Salts with multiple charges
• HPO42- is a protonated salt which behaves as an
  amphoteric substance where the following
  equilibria takes place
• HPO42- D H PO43-
• HPO42- H2O D H2PO4- OH-
• H2O D H OH-
• H HHPO4- Hwater OH-HPO4-
• H PO43- OH- H2PO4-
• H ka3 HPO42-/H kw/H
  HHPO42-/ka2
• Rearrangement of this relation gives
• H (ka2kw ka2ka3 HPO42-)/(ka2
  HPO42-)1/2
• Approximation, if valid, gives
• H (ka2ka3)1/2

7

• Example
• Find the pH of a 0.20 M Na2HPO4 solution. Ka1
  1.1x10-2, ka2 7.5x10-8, ka3 4.8x10-13.
• Solution
• HPO42- is doubly charged so we use ka2 and ka3 as
  the relation above
• H (ka2kw ka2ka3 HPO42-)/(ka2
  HPO42-)1/2
• H (7.5x10-8 10-14 7.5x10-8 4.8x10-13
  0.20)/(7.5x10-8 0.20)1/2 2.0x10-10 M
• pH 9.70

8

• Using the approximated expression we get
• H (7.5x10-8 4.8x10-13)1/2 1.9x10-10 M
• pH 9.72
• This small difference is because ka2kw is not
  very small as compared to the second term and
  thus should be retained.

9

• pH Calculations for Mixtures of Acids
• The key to solving such type of problems is to
  consider the equilibrium of the weak acid and
  consider the strong acid as 100 dissociated as a
  common ion.

10

• Example
• Find the pH of a solution containing 0.10 M HCl
  and 0.10 M HOAc. ka 1.8x10-5
• Solution
• HOAc D H OAc-

11

• Ka (0.10 x) x/(0.10 x)
• Assume 0.1 gtgt x since ka is small
• 1.8x10-5 0.10 x/0.10
• x 1.8x10-5
• Relative error (1.8x10-5/0.10) x100 1.8x10-2
• Therefore H 0.10 1.8x10-5 0.10
• It is clear that all H comes from the strong
  acid since dissociation of the weak acid is
  limited and in presence of strong acid the
  dissociation of the weak acid is further
  suppressed.

12

• Example
• Find the pH of a solution containing 0.10 M HCl
  and 0.10 M H3PO4. Ka1 1.1x10-2, ka2 7.5x10-8,
  ka3 4.8x10-13.
• Solution
• It is clear from the acid dissociation constants
  that ka1gtgtka2 and thus only the first equilibrium
  contributes to the H concentration. Now treat
  the problem as the previous example
• H3PO4 D H2PO4- H

13

• Ka (0.10 x) x/(0.10 x)
• Assume 0.1 gtgt x since ka is small (!!!)
• 1.1x10-2 0.10 x/0.10
• x 1.1x10-2
• Relative error (1.1x10-2/0.10) x100 11
• The assumption is therefore invalid and we have
  to solve the quadratic equation. Result will be
• X 9.2x10-3
• Therefore H 0.10 9.2x10-3 0.11
• pH 0.96

14

• Example
• Find the pH of a solution containing 0.10 M HCl
  and 0.10 M HNO3.
• Solution
• H HHCl HHNO3
• Both are strong acids which are 100 dissociated.
  Therefore, we have
• H 0.10 0.10 0.2
• pH 0.70

15

• In some situations we may have a mixture of two
  weak acids. The procedure for pH calculation of
  such systems can be summarized in three steps
• For each acid, decide whether it is possible to
  neglect dissociations beyond the first
  equilibrium if one or both are polyprotic acids.
• If step 1 succeeds to eliminate equilibria other
  than the first for both acids, compare ka1 values
  for both acids in order to check whether you can
  eliminate either one. You can eliminate the
  dissociation of an acid if ka for the first is
  100 times than ka1 for the second (a factor of
  100 is enough since the acid with the larger ka
  suppresses the dissociation of the other).
• Perform the problem as if you have one acid only
  if step 2 succeeds.

16

• Example
• Find the pH of a solution containing 0.10 M H3PO4
  (ka1 1.1x10-2, ka2 7.5x10-8, ka3 4.8x10-13)
  and 0.10 M HOAc (ka 1.8x10-5).
• Solution
• It is clear for the phosphoric acid that we can
  disregard the second and third equilibria since
  ka1gtgtgt ka2. Therefore we treat the problem as if
  we have the following two equilibria
• H3PO4 D H H2PO4-
• HOAc D H OAc-

17

• Now compare the ka values for both equilibria
• Ka1/ka 1.1x10-2/1.8x10-5 6.1x102
• Therefore the first equilibrium is about 600
  times better than the second. For the moment, let
  us neglect H from the second equilibrium as
  compared to the first.
• Solution for the H is thus simple
• H3PO4 D H2PO4- H

18

• Ka x x/(0.10 x)
• Assume 0.1 gtgt x since ka is small (!!!)
• 1.1x10-2 x2 /0.10
• x 0.033
• Relative error (0.033/0.10) x100 33
• The assumption is therefore invalid and we have
  to solve the quadratic equation. Result will be
• X 0.028

19

• Now let us calculate the H coming from acetic
  acid which is equal to OAc-
• HOAc D H OAc-
• Ka HOAc-/HOAc
• 1.8x10-5 0.028 OAc-/0.10
• OAc- 6.4x10-5 HHOAc
• Relative error (6.4x10-5/0.028) x100 0.23
• Therefore, we are justified to disregard the
  dissociation of acetic acid in presence of
  phosphoric acid. Never calculate the H
  concentration from each acid and add them up.
  This is incorrect.

20
Acid-Base Titrations
21

• In this chapter, we will study titrations of
• 1. Strong acid with strong base
• 2. Strong acid with weak base
• 3. Strong base with weak acid
• 4. Strong acid with polybasic bases
• 5. Strong base with polyprotic acids
• 6. Strong acid with a mixture of two bases
• 7. Strong base with a mixture of two acids.

22

• Acid-Base Indicators
• An acid-base indicator is either a weak acid or
  base which changes color upon changing from one
  chemical form to the other, depending on the pH.
  Indicators are added in a very small amounts in
  order to decrease the titration error. We can
  represent the equilibrium of an indicator as
  follows
• HIn (color 1) D H In- (color 2)

23

• Kin HIn-/HIn
• pkin pH log In-/HIn, or
• pH pkIn log In-/HIn
• Color 1 can be visually observed, in presence of
  color 2, if HIn is at least 10 times In- and
  color 2 can be visually observed, in presence of
  color 1, if In- is at least 10 times HIn.
  Therefore, the final equation can be rewritten
  as
• pH pkIn 1

24

• pH pkIn 1
• This equation is of extreme significance since it
  suggests that
• 1. There is a pH range of two units only where an
  indicator can be used.
• 2. The pkin should be very close to the pH at the
  equivalence point of the titration.
• Therefore, one should look at the pH at the
  equivalence point of the titration in order to
  select the right indicator.

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26

• Once again, the pkin of the indicator should be
  close to the pH of the equivalence point of the
  titration of interest. Look at the following
  titration curve, both indicators have their
  transition ranges on the break of the curve and
  thus either indicator can be used for this
  titration.

27

• However, the following titration curve requires a
  different indicator (the lower one is suitable )

28

• The break on the titration curve is very
  important for accurate determination of the end
  point. As the break becomes steeper and sharper,
  very small excess of a titrant is needed for good
  visualization of the end point. We may think of
  the accuracy of an end point by imagining that
  the indicator starts changing color when the pH
  of the solution touches one side of its range but
  the color is not clear enough unless the pH
  reaches the other side of the range. Look at the
  following titration curve

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30

• The distance between the two arrows represents
  the volume which is needed to undoubtedly observe
  the end point. The color of the indicator starts
  changing at a volume corresponding to the first
  arrow and can be visually seen at a volume
  corresponding to the second arrow. This volume is
  added in excess of the required volume and thus
  an error is obtained corresponding to that volume
  excess.
• A better titration curve is seen below where a
  minimal extra volume is required to visualize the
  end point

31
I have intentionally shown the two arrows
coincide in order to indicate that sharp
inflections or breaks can be very advantageous in
titrations where very low errors should be
expected.
32

• What makes the break sharper?
• This is an important question and the answer is
  rather simple Two reasons
• 1. Concentrations of reactants (analyte and
  titrant) where as concentration increase,
  sharpness of the break increase.
• 2. The dissociation constant where as the
  dissociation constant increases, the sharpness of
  the break increases. This suggests that strong
  acids and bases are expected to have sharper
  breaks while weak acids are expected to have
  diffused breaks.

33

• Strong acids can be titrated with strong bases
  leading to formation of salts and water. Remember
  that the salt formed from a strong acid and
  strong base is a combination between the weak
  conjugate base of the strong acid and the weak
  conjugate acid of the strong base. Both
  conjugates are weak and thus do not react with
  water which means a H 10-7 M and pH7 at the
  equivalence point. Let us now look at an example
  for such a titration

34

• Example
• Calculate the pH of a 50 mL 0f 0.10 M HCl after
  addition of 0, 20, 40, 50, 80, and 100 mL of 0.1
  M NaOH.
• Solution
• After addition of 0 mL NaOH we only have HCl.
• H 0.10 M , same as HCl concentration since
  HCl is a strong acid
• pH 1.0

35

• After addition of 20 mL NaOH
• mmol H left 0.150 0.120 3.0
• H 3.0/70 0.043
• pH 1.34
• After addition of 40 mL NaOH
• mmol H left 0.150 0.140 1.0
• H 1.0/80 0.0125
• pH 1.90
• Same steps are used for calculation of pH at any
  point before equivalence.

36

• After addition of 50 mL NaOH
• mmol H left 0.150 0.150 0
• Therefore, this is the equivalence point and the
  H will only be produced from dissociation of
  water
• H2O D H OH-
• H OH- 10-7 M
• pH 7

37

• After addition of 80 mL NaOH
• mmol OH- excess 0.180 0.150 3.0
• OH- 3.0/130 0.023
• pOH 1.63
• pH 14 1.63 12.37
• The same procedure is used for calculation of any
  point after equivalence is reached.

38

• Titration of a Weak Acid with a Strong Base
• Weak acids could be titrated against strong bases
  where the following points should be remembered
• 1. Before addition of any base, we have a
  solution of the weak acid and calculation of the
  pH of the weak acid should be performed as in
  previous sections.
• 2. After starting addition of the strong base to
  the weak acid, the salt of the weak acid is
  formed. Therefore, a buffer solution results and
  you should consult previous lectures to find out
  how the pH is calculated for buffers.

39

• 3. At the equivalence point, the amount of strong
  base is exactly equivalent to the weak acid and
  thus there will be 100 conversion of the acid to
  its salt. The problem now is to calculate the pH
  of the salt solution.
• 4. After the equivalence point, we would have a
  solution of the salt with excess strong base. The
  presence of the excess base suppresses the
  dissociation of the salt in water and the pH of
  the solution comes from the excess base only.
• Now, let us apply the abovementioned concepts on
  an actual titration of a weak acid with a strong
  base.

40

• Example
• Find the pH of a 50 mL solution of 0.10 M HOAc
  (ka 1.75x10-5) after addition of 0, 10, 25, 50,
  60 and 100 mL of 0.10 M NaOH.
• Solution
• After addition of 0 mL NaOH
• we have the acid only where
• HOAc D H OAc-

41

• Ka HOAc-/HOAc
• Ka x x / (0.10 x)
• Ka is very small. Assume 0.10 gtgt x
• 1.7510-5 x2/0.10
• x 1.3x10-3
• Relative error (1.3x10-3/0.10) x 100 1.3
• The assumption is valid and the H 1.3x10-3 M
• pH 2.88

42

• 2. After addition of 10 mL NaOH
• initial mmol HOAc 0.10 x 50 5.0
• mmol NaOH added 0.10 x 10 1.0
• mmol HOAc left 5.0 1.0 4.0
• HOAc 4.0/60 M
• mmol OAc- formed 1.0
• OAc- 1.0/60 M
• HOAc D H OAc-

43

• Ka x (1.0/60 x)/ (4.0/60 x)
• Assume 21.0/60 gtgt x
• 1.75x10-5 x (1.0/60)/4.0/60
• x 1.75x10-5 x 1.0/4.0
• x 7.0 x10-5
• Relative error 7.0 x10-5/(1.0/60) x 100
  0.42
• The assumption is valid
• pH 4.15

44

• 3. After addition of 25 mL NaOH
• initial mmol HOAc 0.10 x 50 5.0
• mmol NaOH added 0.10 x 25 2.5
• mmol HOAc left 5.0 2.5 2.5
• HOAc 2.5/75 M
• mmol OAc- formed 2.5
• OAc- 2.5/75 M
• HOAc H OAc-

45

• Ka x (2.5/75 x)/ (2.5/75 x)
• Assume 2.5/75 gtgt x
• 1.75x10-5 x (2.5/75)/(2.5/75)
• x 1.75x10-5
• Relative error 1.75x10-5/(2.5/75) x 100
  0.052
• The assumption is valid
• pH 4.76
• Note here that at 25 mL NaOH (half the way to
  equivalence point) pH pka . It is therefore
  possible to find the ka of a weak monoprotic
  acid by calculating half the volume of strong
  base needed to reach the equivalence point. At
  this volume ka H.

46

• 4. After addition of 50 mL NaOH
• initial mmol HOAc 0.10 x 50 5.0
• mmol NaOH added 0.10 x 50 5.0
• mmol HOAc left 5.0 5.0 0
• This is the equivalence point
• mmol OAc- formed 5.0
• OAc- 5.0/100 0.05 M
• Now the solution we have is a 0.05 M acetate
• OAc- H2O D HOAc OH-

47

• Kb kw/ka
• Kb 10-14/1.75x10-5 5.7x10-10
• Kb HOAcOH-/OAc-
• Kb x x/(0.05 x)
• Kb is very small and we can fairly assume that
  0.05gtgtx
• 5.7x10-10 x2/0.05
• x 5.33 x 10-6
• Relative error (7.6x10-6/0.05) x100 0.011
• The assumption is valid.
• OH- 5.33x10-6 M
• pOH 5.27
• pH 14 5.27 8.73

48

• 5. After addition of 60 mL NaOH
• Initial mmol of HOAc 0.10 x 50 5.0
• mmol NaOH added 0.10 x 60 6.0
• mmol NaOH excess 6.0 5.0 1.0
• OH- 1.0/110
• pOH 2.04
• pH 14 2.04 11.96

• 6. After addition of 100 mL NaOH
• Initial mmol of HOAc 0.10 x 50 5.0
• mmol NaOH added 0.10 x 100 10.0
• mmol NaOH excess 10.0 5.0 5.0
• OH- 5.0/150
• pOH 1.48
• pH 14 1.48 12.52